Question

The heat of vaporization of a certain refrigerant is 162 J/g. Calculate the number of
kilograms of refrigerant that must be evaporated to freeze a tray of 16 one-ounce
(1 oz = 28 g) ice cubes starting with the water at 16°C.
Heat capacity (water) = 4.184 J/g °C
ΔHfusion (water) = 333 J/g

Answer 1

Answer:

We need 1.106 kilograms of refrigerant to freeze 16 one-ounce ice cubes.

Explanation:

According to the statement, the refrigerant must receive the heat released from water to create ice cubes. If we know that there is no heat nor mass interactions with surroundings and freezing process is steady, then we get the following model from First Law of Thermodynamics:

$U_{r, 1} - U_{r, 2} + U_{w, 1}-U_{w,2} = 0$ (Eq. 1)

Where:

$U_{r,1}$, $U_{r,2}$ - Initial and final internal energies of the refrigerant, measured in joules.

$U_{w,1}$, $U_{w,2}$ - Initial and final internal energies of water, measured in joules.

Please notice that refrigerant is only vaporized (latent), whereas water is cooled down and freezed (sensible + latent). Then, we expand the equation above by definitions of sensible and latent heat:

$m_{r}\cdot h_{v,r} + m_{w}\cdot [c_{w}\cdot (T_{w,f}-T_{w,o})-h_{f,w}] = 0$ (Eq. 2)

Where:

$m_{r}$, $m_{w}$ - Masses of the refrigerant and water, measured in grams.

$h_{v,r}$ - Latent heat of vaporization of the refrigerant, measured in joules per gram.

$h_{f,w}$ - Latent heat of fusion of the water, measured in joules per gram.

$T_{w,o}$, $T_{w,f}$ - Initial and final temperatures of water, measured in degrees Celsius.

Then we clear the mass of the refrigerant within:

$m_{r} = -\frac{m_{w}\cdot [c_{w}\cdot (T_{w,f}-T_{w,o})-h_{f,w}]}{h_{v,r}}$

If we know that $m_{w} = 448\,g$, $c_{w} = 4.184\,\frac{J}{g\cdot ^{\circ}C}$, $T_{w, o} = 16\,^{\circ}C$, $T_{w,f} = 0\,^{\circ}C$, $h_{f,w} = 333\,\frac{J}{g}$ and $h_{v,r} = 162\,\frac{J}{g}$, then the mass of the refrigerant is:

$m_{r} = -\frac{(448\,g)\cdot \left[\left(4.184\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (0\,^{\circ}C-16\,^{\circ}C)-333\,\frac{J}{g} \right]}{162\,\frac{J}{g} }$

$m_{r} = 1106.018\,g$

If we know that a kilogram equals 1000 grams, then we need 1.106 kilograms of refrigerant to freeze 16 one-ounce ice cubes.